今天学习了一下算法导论上的快速傅里叶变换,并写了个模板,留作今后使用吧~~~
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <complex>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long Int64;
const int MAXN = 200010;
const double PI = acos(-1.0);
typedef complex<double> Complex;
char str1[MAXN], str2[MAXN];
Complex A[MAXN], B[MAXN];
int str_len1, str_len2, len;
int ans[MAXN];
void bulid(Complex _P[], Complex P[], int n, int m, int curr, int &cnt)
{
if(m == n)
{
_P[curr] = P[cnt++];
}
else
{
bulid(_P, P, n, m * 2, curr, cnt);
bulid(_P, P, n, m * 2, curr + m, cnt);
}
return ;
}
/*
*P 需要进行DFT变换的数据
*n 数据长度(2的整数次幂)
*oper 1(正变换), -1(负变换)
*/
void FFT(Complex P[], int n, int oper)
{
static Complex _P[MAXN];
int cnt = 0;
bulid(_P, P, n, 1, 0, cnt);
copy(_P, _P + n, P);
for(int d = 0; (1 << d) < n; d++)
{
int m = 1 << d;
int m2 = m * 2;
double p0 = PI / m * oper;
Complex unit_p0 = Complex(cos(p0), sin(p0));
for(int i = 0; i < n; i += m2)
{
Complex unit = 1;
for(int j = 0; j < m; ++j)
{
Complex &P1 = P[i + j + m], &P2 = P[i + j];
Complex t = unit * P1;
P1 = P2 - t;
P2 = P2 + t;
unit = unit * unit_p0;
}
}
}
if(oper == -1)
for(int i = 0; i < n; ++i)
P[i] = Complex(P[i].real()/n, P[i].imag());
return ;
}
int main()
{
//freopen("aa.in", "r", stdin);
//freopen("bb.out", "w", stdout);
while(scanf("%s %s", str1, str2) != EOF)
{
memset(ans, 0, sizeof(ans));
str_len1 = strlen(str1);
str_len2 = strlen(str2);
int len = 1;
while(len < (2*str_len1) || len < (2*str_len2))
len <<= 1;
for(int i = 0; i < str_len1; ++i)
{
A[i] = Complex(static_cast<double>(str1[str_len1-i-1]-'0'), 0.0);
}
for(int i = str_len1; i < len; ++i)
{
A[i] = Complex(0.0, 0.0);
}
for(int i = 0; i < str_len2; ++i)
{
B[i] = Complex(static_cast<double>(str2[str_len2-i-1]-'0'), 0.0);
}
for(int i = str_len2; i < len; ++i)
{
B[i] = Complex(0.0, 0.0);
}
FFT(A, len, 1); FFT(B, len, 1);
for(int i = 0; i < len; ++i)
{
A[i] = A[i] * B[i];
}
FFT(A, len, -1);
for(int i = 0; i < len; ++i)
{
ans[i] = static_cast<int>(A[i].real() + 0.5);
}
for(int i = 0; i < len; ++i)
{
ans[i + 1] += ans[i] / 10;
ans[i] %= 10;
}
len = str_len1 + str_len2;
while(!ans[len] && len > 0)
len--;
for(int i = len; i >= 0; --i)
putchar((ans[i]+'0'));
putchar('\n');
}
return 0;
}
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <complex>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 200010;
const double PI = acos(-1.0);
struct Complex
{
double r, i;
Complex(double _r = 0.0, double _i = 0.0)
{
this->r = _r;
this->i = _i;
}
Complex operator+ (const Complex &b)
{
return Complex(r + b.r, i + b.i);
}
Complex operator- (const Complex &b)
{
return Complex(r - b.r, i - b.i);
}
Complex operator* (const Complex &b)
{
return Complex(r * b.r - i * b.i, r * b.i + i * b.r);
}
};
/*
* 进行FFT和IFFT前的反转变换。
* 位置i和 (i二进制反转后位置)互换
* len必须去2的幂
*/
void change(Complex y[], int len)
{
int i, j, k;
for(i = 1, j = len/2; i < len - 1; ++i)
{
if(i < j)
swap(y[i], y[j]);
k = len / 2;
while(j >= k)
{
j -= k;
k /= 2;
}
if(j < k)
j += k;
}
}
/*
*FTT求解
*len必须为2^k形式
*on==1时DFT,on==-1时是DFT
*/
void FFT(Complex y[], int len, int on)
{
change(y, len);
for(int h = 2; h <= len; h <<= 1)
{
Complex wn(cos(-on*2*PI/h), sin(-on*2*PI/h));
for(int j = 0; j < len; j += h)
{
Complex w(1, 0);
for(int k = j; k < j + h/2; ++k)
{
Complex u = y[k];
Complex t = w * y[k+h/2];
y[k] = u + t;
y[k+h/2] = u - t;
w = w * wn;
}
}
}
if(on == -1)
for(int i = 0; i < len; ++i)
y[i].r /= len;
return ;
}
Complex x1[MAXN], x2[MAXN];
char str1[MAXN], str2[MAXN];
int len1, len2, len;
int ans[MAXN];
int main()
{
//freopen("aa.in", "r", stdin);
//freopen("bb.out", "w", stdout);
while(scanf("%s %s", str1, str2) != EOF)
{
memset(ans, 0, sizeof(ans));
len1 = strlen(str1), len2 = strlen(str2);
len = 1;
while(len < 2 * len1 || len < 2 * len2)
len <<= 1;
for(int i = 0; i < len1; ++i)
x1[i] = Complex((str1[len1-i-1]-'0'), 0.0);
for(int i = len1; i < len; ++i)
x1[i] = Complex(0.0, 0.0);
for(int i = 0; i < len2; ++i)
x2[i] = Complex((str2[len2-i-1]-'0'), 0.0);
for(int i = len2; i < len; ++i)
x2[i] = Complex(0.0, 0.0);
FFT(x1, len, 1); FFT(x2, len, 1);
for(int i = 0; i < len; ++i)
x1[i] = x1[i] * x2[i];
FFT(x1, len, -1);
for(int i = 0; i < len; ++i)
{
//cout << "Test: " << x1[i].r << endl;
ans[i] = (int)(x1[i].r + 0.5);
}
for(int i = 0; i < len; ++i)
{
ans[i+1] += ans[i] / 10;
ans[i] %= 10;
}
len = len1 + len2 - 1;
while(ans[len] <= 0 && len > 0)
len--;
for(int i = len; i >= 0; --i)
putchar((ans[i]+'0'));
putchar('\n');
}
return 0;
}